By Madhu Sudan

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We still have to deal with the issue of small values of jRj. But note that if jRj is small, it is in particular nite, and hence R is in fact a eld. Now given f 2 R x; y], we work with a nite extension R0 of R such that jR0j > 4d2 and obtain a non-trivial irreducible factor (if any) g0 2 R0 x; y] of f 2 R0 x; y] using the above corollary. We now need to recover a factor g 2 R x; y] using g0 2 R0 x; y]. 3 Let g0 ; f be as above. Let g 2 R x; y] be the minimal polynomial such that g0 (x; y)jg(x; y) (in R0 x; y]).

Notice that g0 does not necessarily correspond to a factor of f. However, g0 does divide (modulo y) an irreducible factor of f. In other words, there exist polynomials g(x; y); h(x; y) such that f(x; y) = g(x; y)h(x; y), g(x; y) is irreducible and g(x; y) = g0 (x; y)l0 (x; y) (mod y) and h0 (x; y) = l0 (x; y)h(x; y) (mod y). 2). 2). Proof Notice rst g satis es the degree conditions and the condition g 6= 0. It su ce to show that there exists lk such that g(x; y) = gk (x; y)lk (x; y) (mod y2k ).

15). Since f~i (x; t) = fa^i ;^b (x; t), the lemma follows. 1 Factoring Univariate Polynomials over Integers In this lecture, we shall discuss factorisation of univariate polynomials over integers. Factoring polynomials over the rationals can be reduced to this case by a clearing of denominators. This factorisation can be further extended to factoring multivariate P polynomials as discussed in the previous 3 lectures. Given a polynomial f(x) = ni=0 ai xi 2 Zx];(an 6= 0), factoring f involves nding irreducible polynomials f1 ; f2; : : :fk 2 Zx] and c 2 Zsuch that f(x) = cf1 (x)f2 (x) : : :fk (x).